Kvant Physics Problem 2

Two point masses $m_1$ and $m_2$, with units kilograms, lie on a horizontal frictionless plane and are connected by a light spring with stiffness $c$ measured in $\mathrm{N,m^{-1}}$.

kvantphysics

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m25s
Source on kvant.digital

Problem

On a horizontal plane lie two balls with masses $m_1$ and $m_2$, connected by a spring with stiffness $c$. The plane is frictionless. The balls are displaced, compressing the spring, and then released simultaneously. Determine the periods of the resulting oscillations of the balls.

Setup and Assumptions

Two point masses $m_1$ and $m_2$, with units kilograms, lie on a horizontal frictionless plane and are connected by a light spring with stiffness $c$ measured in $\mathrm{N,m^{-1}}$. The natural length of the spring is denoted by $\ell_0$ with units meters. The motion is restricted to one dimension along the line joining the masses.

The system is released from rest after the spring has been compressed, so the initial velocities of both masses are zero. The spring is assumed ideal, obeying Hooke’s law at all deformations, and its mass is neglected compared to $m_1$ and $m_2$. No external horizontal forces act on the system, so the center of mass moves uniformly.

The unknown quantity is the period of oscillation of each mass about the equilibrium configuration.

Physical Principles

Newton’s second law is applied to each mass in the form

$m_1 \ddot{x}_1 = F_1, \qquad m_2 \ddot{x}_2 = F_2,$

where $x_1(t)$ and $x_2(t)$ are the coordinates of the masses.

The spring force follows Hooke’s law,

$F = -c(\Delta \ell),$

where $\Delta \ell = (x_2 - x_1 - \ell_0)$ is the deviation from the natural length.

The center of mass coordinate is defined as

$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2},$

and in the absence of external forces it satisfies $\ddot{X} = 0$.

The relative motion is described using a reduced mass formulation,

$\mu = \frac{m_1 m_2}{m_1 + m_2},$

which converts the two-body interaction into an equivalent one-body oscillation problem.

Derivation

The equations of motion for the two masses are

$m_1 \ddot{x}_1 = c(x_2 - x_1 - \ell_0),$

$m_2 \ddot{x}_2 = -c(x_2 - x_1 - \ell_0).$

Subtracting the first equation from the second eliminates the center of mass motion. Introducing the relative coordinate

$y = x_2 - x_1 - \ell_0,$

its second derivative is

$\ddot{y} = \ddot{x}_2 - \ddot{x}_1.$

Using the equations of motion,

$\ddot{y} = -\frac{c}{m_2}(x_2 - x_1 - \ell_0) - \frac{c}{m_1}(x_2 - x_1 - \ell_0),$

which becomes

$\ddot{y} = -c\left(\frac{1}{m_1} + \frac{1}{m_2}\right)y.$

The coefficient is rewritten using the reduced mass,

$\frac{1}{\mu} = \frac{1}{m_1} + \frac{1}{m_2},$

so the equation of motion takes the standard harmonic form

$\ddot{y} + \frac{c}{\mu} y = 0.$

The angular frequency is

$\omega = \sqrt{\frac{c}{\mu}} = \sqrt{c\left(\frac{1}{m_1} + \frac{1}{m_2}\right)}.$

The period of oscillation is

$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\mu}{c}}.$

Both masses execute harmonic motion with this same angular frequency, since each coordinate is a linear combination of the center of mass motion (uniform) and the relative oscillation (harmonic).

Result

The reduced mass is

$\mu = \frac{m_1 m_2}{m_1 + m_2} \ \mathrm{kg}.$

The oscillation period of each mass is

$T = 2\pi \sqrt{\frac{m_1 m_2}{(m_1 + m_2)c}} \ \mathrm{s}.$

Thus,

$\boxed{T_1 = T_2 = 2\pi \sqrt{\frac{m_1 m_2}{(m_1 + m_2)c}} \ \mathrm{s}}.$

Sanity Checks

The expression has dimensions of time since $c$ has units $\mathrm{N,m^{-1}} = \mathrm{kg,s^{-2}}$, so $\mu/c$ has units $\mathrm{s^2}$ and its square root has units $\mathrm{s}$.

In the limit $m_1 \gg m_2$, the reduced mass becomes $\mu \approx m_2$, and the period reduces to $T \approx 2\pi \sqrt{m_2/c}$, corresponding to the lighter mass oscillating against an effectively fixed support provided by the heavier mass.

In the symmetric case $m_1 = m_2 = m$, the reduced mass becomes $\mu = m/2$, giving $T = 2\pi \sqrt{m/(2c)}$, which is shorter than the single-mass spring period due to both masses sharing the inertia.

A sign error would most likely appear when forming the relative coordinate equation, where the restoring force must enter with opposite signs for the two masses; this would incorrectly change the sign of the effective spring term and destroy the oscillatory solution.